Matrices
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Elementary Transformation
Rules for Elementary Transformation :
In Elementary Transformation, you can
1) Interchange any two rows or columns
2) Add or subtract any two rows or columns
3) Divide or multiply any row or column with a constant (and NOT by any other row or column)
Example :
\[ \text{Let} \hspace{10pt} \text{A} = \begin{bmatrix} 3 & 2 & 1 \\ 2 & 5 & 7 \\ 1 & 9 & 10 \end{bmatrix}, \text{then} \] \[ \text{1)} R_1 \longleftrightarrow R_2 \hspace{5pt} \sim \begin{bmatrix} 2 & 5 & 7 \\ 3 & 2 & 1 \\ 1 & 9 & 10 \end{bmatrix} \] \[ \text{2)} R_1 \rightarrow R_1 - R_2 \hspace{5pt} \sim \begin{bmatrix} 1 & -3 & -6\\ 2 & 5 & 7\\ 1 & 9 & 10 \end{bmatrix} \] \[ \text{3)} R_1 \rightarrow 3 \times R_1 \hspace{5pt} \sim \begin{bmatrix} 9 & 6 & 3\\ 2 & 5 & 7\\ 1 & 9 & 10 \end{bmatrix} \] -
Rank of a Matrix
Rank :
Rank r means there exists at least one r-rowed square matrix with non-vanishing determinant.\(\Rightarrow\) every (r+1) or more rowed matrices have vanishing determinants.
Also, rank of a matrix is the largest order of a non-zero minor of the matrix(i.e. determinant not equal to 0).
Minor \(\Rightarrow\) Part of a matrix \(\Rightarrow\) always square ordered
Example: Let A is a matrix of order \(4 \times 3\) \[ A = \begin{bmatrix} 3 & 1 & 4\\ 0 & 5 & 8\\ -3 & 4 & 4\\ 1 & 2 & 4 \end{bmatrix} \] Then 4 minors are possible by dropping one row each time:
a) Leaving \(R_1 \longrightarrow \begin{bmatrix} 0 & 5 & 8\\ -3 & 4 & 4\\ 1 & 2 & 4 \end{bmatrix} \)
b) Leaving \(R_2 \longrightarrow \begin{bmatrix} 3 & 1 & 4\\ -3 & 4 & 4\\ 1 & 2 & 4 \end{bmatrix} \)
c) Leaving \(R_3 \longrightarrow \begin{bmatrix} 3 & 1 & 4\\ 0 & 5 & 8\\ 1 & 2 & 4 \end{bmatrix} \)
d) Leaving \(R_4 \longrightarrow \begin{bmatrix} 3 & 1 & 4\\ 0 & 5 & 8\\ -3 & 4 & 4 \end{bmatrix} \)
Properties :
1) Rank of two equivalent matrices is always same.
2) Rank of \(A\) and \(A^T\) is same.
3) Rank of a null matrix is same.
4) For a rectangular matrix \(A\) of order m \(\times\) n, rank of \(A \leq min(m,n)\).
5) For a \(n\)-square matrix, if rank = \(n\), then \(|A| \neq 0\) i.e. \(A\) is non-singular.
6) For any square matrix, if rank < \(n\), then \(|A| = 0\) i.e. \(A\) is singular.
Sample Problem :
Find the rank of the matrix \( \begin{bmatrix} 3 & 1 & 4\\ 0 & 5 & 8\\ -3 & 4 & 4\\ 1 & 2 & 4 \end{bmatrix} \)
Solution:
Echelon Form: The approach will be: (a) reduce the diagonal elements to 1 and the rest of the elements to 0 with the help of Elementary Transformations, (b) after that, the number of non-zero rows will be the rank of the matrix. So starting with first column, reduce first element of \(R_1\) to 1 and with the help of that element reduce all the rest of the elements of that row(or column) to 0.
\[ R_1 \longleftrightarrow R_4 \] \[ \sim \begin{bmatrix} 1 & 2 & 4\\ 0 & 5 & 8\\ -3 & 4 & 4\\ 3 & 1 & 4 \end{bmatrix} \] \[ R_3 \rightarrow R_3 + 3R_1, \hspace{10pt} R_4 \rightarrow R_4 - 3R_1 \] \[ \sim \begin{bmatrix} 1 & 2 & 4\\ 0 & 5 & 8\\ 0 & 10 & 16\\ 0 & -5 & -8 \end{bmatrix} \] Now, as the rest of the elements are 0, move on to the next column and perform same steps \[ C_2 \rightarrow C_2 - 2C_1, \hspace{10pt} C_3 \rightarrow C_3 - 4C_1 \] \[ \sim \begin{bmatrix} 1 & 0 & 0\\ 0 & 5 & 8\\ 0 & 10 & 16\\ 0 & -5 & -8 \end{bmatrix} \] \[ C_2 \rightarrow \frac{C_2}{5} \] \[ \sim \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 8\\ 0 & 2 & 16\\ 0 & -1 & -8 \end{bmatrix} \] \[ C_3 \rightarrow C_3 - 8C_2 \] \[ \sim \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \] Since, it has 2 non-zero rows, Rank \(\hspace{10pt}R = 2\)
Normal Form:
Any matrix can be reduced to following 4 forms: \[ a) \hspace{20pt} \begin{bmatrix} I_3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \] \[ b) \hspace{20pt} \begin{bmatrix} I_3\\ 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \hspace{10pt} \text{(only zero rows, no zero columns)} \] \[ c) \hspace{20pt} \begin{bmatrix} I_3 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{bmatrix} \hspace{10pt} \text{(only zero columns, no zero rows)} \] \[ d) \hspace{20pt} \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} I_3 & 0\\ 0 & 0 \end{bmatrix} \hspace{10pt} \text{(zero rows and columns)} \]
Sample Problem:
Find normal form of matrix \( \begin{bmatrix} 2 & 1 & -3 & -6\\ 3 & -3 & 1 & 2\\ 1 & 1 & 1 & 2 \end{bmatrix} \)
Solution:
\[ R_1 \longleftrightarrow R_3 \] \[ \sim \begin{bmatrix} 1 & 1 & 1 & 2\\ 3 & -3 & 1 &2\\ 2 & 1 & -3 & -6 \end{bmatrix} \] \[ R_2 \rightarrow R_2 - 3R_1, \hspace{10pt} R_3 \rightarrow R_3 - 2R_1 \] \[ \sim \begin{bmatrix} 1 & 1 & 1 & 2\\ 0 & -6 & -2 & -4\\ 0 & -1 & -5 & -10 \end{bmatrix} \] \[ C_2 \rightarrow C_2 - C_1, \hspace{10pt} C_3 \rightarrow C_3 - C_1, \hspace{10pt} C_4 \rightarrow C_4 - 2C_1 \] \[ \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & -6 & -2 & -4\\ 0 & -1 & -5 & -10 \end{bmatrix} \] \[ R_2 \longleftrightarrow R_3 \] \[ \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & -5 & -10\\ 0 & -6 & -2 & -4 \end{bmatrix} \] \[ R_2 \rightarrow -R_2 \] \[ \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 5 & 10\\ 0 & -6 & -2 & -4 \end{bmatrix} \] \[ R_3 \rightarrow R_3 + 6R_2 \] \[ \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 5 & 10\\ 0 & 0 & 28 & 56 \end{bmatrix} \] \[ C_3 \rightarrow C_3 - 5C_2, \hspace{10pt} C_4 \rightarrow C_4 - 10C_2 \] \[ \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 28 & 56 \end{bmatrix} \] \[ R_3 \rightarrow \frac{R_3}{28} \] \[ \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 2 \end{bmatrix} \] \[ C_4 \rightarrow C_4 - 2C_3 \] \[ \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} I_3 & 0 \end{bmatrix} \] \[ \text{Also, 3 non-zero rows so rank } \boxed{R = 3} \]
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Solution of System of Linear Equations
Linear Equations:
Equations in which degree of the variables is 1.
A set of Linear Equations can be written in matrix form as \(A, X, \) and \( B\) where \(A = \text{coefficient matrix, }X = \text{variables matrix, and }B = \text{constant matrix}\)
So to find solution, first we take augmented matrix of \(A\) and \(B\) i.e. [\(A:B\)] which is matrix \(A\) and \(B\) written in one matrix seperated by a colon (:).
Then we find rank of \(A\) and [\(A:B\)] after which arise two situations:
1) \(r(A) = r(A:B) \) \(\Rightarrow Consistent\) : again two situations:
\(\hspace{20pt}\)a) \(r(A) = r(A:B) = n\) : (where n is number of variables) the equations have Unique solution
\(\hspace{20pt}\)b) \(r(A) = r(A:B) < n\) : the equations have Infinite solutions
in both the situations, solution is found out through equation : \(AX = B\)
2) \(r(A) \neq r(A:B) \) \(\Rightarrow Inconsistent\) : no solutions
Sample Problem 1:
Check consistency and solve: \( \hspace{20pt} \begin{align*} x + 2y + z &= 3\\ 2x + 3y + 2z &= 5\\ 3x - 5y + 5z &= 2\\ 3x + 9y - z &= 4 \end{align*} \)
Solution :
\[A = \begin{bmatrix} 1 & 2 & 1\\ 2 & 3 & 2\\ 3 & -5 & 5\\ 3 & 9 & -1 \end{bmatrix}, \hspace{10pt} X = \begin{bmatrix} x\\ y\\ z \end{bmatrix}, \hspace{10pt} B = \begin{bmatrix} 3\\ 5\\ 2\\ 4 \end{bmatrix} \] \[ A:B = \hspace{20pt} \begin{bmatrix} 1 & 2 & 1 & : &3\\ 2 & 3 & 2 & : &5\\ 3 & -5 & 5 & : &2\\ 3 & 9 & -1 & : &4 \end{bmatrix} \] \[ R_2 \rightarrow R_2 - 2R_1, \hspace{10pt} R_3 \rightarrow R_3 - 3R_1, \hspace{10pt} R_4 \rightarrow R_4 - 3R_1 \] \[ \sim \begin{bmatrix} 1 & 2 & 1 &: & 3\\ 0 & -1 & 0 &: &-1\\ 0 & -11 & 2 &: &-7\\ 0 & 3 & -4 &: &-5 \end{bmatrix} \] \[ R_2 \rightarrow (-R_2) \] \[ \sim \begin{bmatrix} 1 & 2 & 1 &: &3\\ 0 & 1 & 0 &: &1\\ 0 & -11 & 2 &: &-7\\ 0 & 3 & -4 &: &-5 \end{bmatrix} \] \[ R_3 \rightarrow R_3 + 11R_2, \hspace{10pt} R_4 \rightarrow R_4 - 3R_1 \] \[ \sim \begin{bmatrix} 1 & 2 & 1 &: &3\\ 0 & 1 & 0 &: &1\\ 0 & 0 & 2 &: &4\\ 0 & 0 & -4 &: &-8 \end{bmatrix} \] \[ R_4 \rightarrow R_4 + 2R_3 \] \[ \sim \begin{bmatrix} 1 & 2 & 1 &: &3\\ 0 & 1 & 0 &: &1\\ 0 & 0 & 2 &: &4\\ 0 & 0 & 0 &: &0 \end{bmatrix} \] Rank \(A = 3\), Rank \(A:B = 3 \Rightarrow\) Equations are Consistent
Since \(r(A) = r(A:B) = 3 = n \) (number of variables) \(\Rightarrow \)Unique solution
For solution, \(AX = B\) \[ \begin{bmatrix} 1 & 2 & 1\\ 0 & 1 & 0\\ 0 & 0 & 2\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 3\\ 1\\ 4\\ 0 \end{bmatrix} \] \[ \begin{align*} x + 2y + z = 3 \hspace{15pt} ,\\ \hspace{20pt} &\boxed{y = 1} \hspace{15pt} \\ 2z = 4 \hspace{10pt} \Rightarrow &\boxed{z = 2} \hspace{10pt} \\ x + 2 + 2 = 3 \hspace{15pt} \Rightarrow &\boxed{x = -1} \end{align*} \]
Sample Problem 2 :
Test the consistency of following equations and solve them if consistent: \[ \begin{align*} x + 2y + 3t &= 2\\ 2x + y + z + t &= -4\\ 4x -3y + z + 7t &= 8 \end{align*} \]
Solution:
\(A = \begin{bmatrix} 1 & 2 & 0 & 3\\ 2 & 1 & 1 & 1\\ 4 & -3 & 1 & 7 \end{bmatrix} \hspace{10pt} X = \begin{bmatrix} x\\ y\\ z\\ t \end{bmatrix} \hspace{10pt} B = \begin{bmatrix} 2\\ -4\\ 8 \end{bmatrix} \) \[ A : B = \begin{bmatrix} 1 &2 &0 &3 &: &2\\ 2 &1 &1 &1 &: &-4\\ 4 &3 &1 &7 &: &8 \end{bmatrix} \] \[ R_2 \rightarrow R_2 - 2R_1, \hspace{10pt} R_3 \rightarrow R_3 - 4R_1 \] \[ \sim \begin{bmatrix} 1 &2 &0 &3 &: &2\\ 0 &-3 &1 &-5 &: &-8\\ 0 &-11 &1 &-5 &: &0 \end{bmatrix} \] \[ R_2 \rightarrow \frac{R_2}{-3} \] \[ \sim \begin{bmatrix} 1 &2 &0 &3 &: &2\\ 0 &1 &\frac{-1}{3} &\frac{5}{3} &: &\frac{8}{3}\\ 0 &-11 &1 &-5 &: &0 \end{bmatrix} \] \[ R_3 \rightarrow R_3 + 11R_2 \] \[ \sim \begin{bmatrix} 1 &2 &0 &3 &: &2\\ 0 &1 &\frac{-1}{3} &\frac{5}{3} &: &\frac{8}{3}\\ 0 &0 &\frac{-8}{3} &\frac{40}{3} &: &\frac{88}{3} \end{bmatrix} \] \[ R_3 \rightarrow \frac{-3}{8}R_3 \] \[ \sim \begin{bmatrix} 1 &2 &0 &3 &: &2\\ 0 &1 &\frac{-1}{3} &\frac{5}{3} &: &\frac{8}{3}\\ 0 &0 &1 &-5 &: &-11 \end{bmatrix} \] \[\begin{align*} r(A) = r(A:B) = 3 &\Rightarrow Consistent\\ r(A) = r(A:B) < \text{number of variables } &\Rightarrow Infinite solutions\\ \text{For solution} : AX = B \end{align*} \] \[ \begin{bmatrix} 1 &2 &0 &3\\ 0 &1 &\frac{-1}{3} &\frac{5}{3}\\ 0 &0 &1 &-5 \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ t \end{bmatrix} = \begin{bmatrix} 2\\ \frac{8}{3}\\ -11 \end{bmatrix} \] \[ \begin{align*} x + 2y + 3t &= 2 \hspace{10pt} \longrightarrow (1) \\ y - \frac{z}{3} + 5\frac{t}{3} &= \frac{8}{3} \\ 3y - z + 5t &= 8 \hspace{10pt} \longrightarrow (2) \\ z - 5t &= -11 \hspace{10pt} \longrightarrow (3) \\ \end{align*} \] \( \begin{align*} Let \hspace{5pt}t = k \\ \text{Put value of }t \text{ in } (3) \end{align*} \) \[ \begin{align*} z - 5k = -11\\ \hspace{25pt} \boxed{z = 5k - 11} \end{align*} \] \[ \begin{align*} \text{In equation} (2), \hspace{10pt} 3y - \cancel{5k} + 11 + \cancel{5k} &= 8\\ 3y = -3 \hspace{10pt} \Rightarrow \boxed{y = -1} \end{align*} \] \[\begin{align*} \text{In equation }(1), \hspace{10pt} x - 2 + 3k &= 2\\ \boxed{x = 4 -3k} \end{align*} \]
Sample Problem 3:
Find values of \(\lambda\) and \(\mu\) for which given system of equations has (a) Infinite solution, (b) No solutions, (c) Unique solution \[ \begin{align*} x + y + z &= 6\\ x + 2y + 3z &= 10\\ x + 2y + \lambda &= \mu \end{align*} \]
Solution:
\[ A = \begin{bmatrix} 1 &1 &1\\ 1 &2 &3\\ 1 &2 &\lambda \end{bmatrix}, X = \begin{bmatrix} x\\ y\\ z \end{bmatrix} B = \begin{bmatrix} 6\\ 10\\ \mu \end{bmatrix} \] \[ A : B = \begin{bmatrix} 1 &1 &1 &: &6\\ 1 &2 &3 &: &10\\ 1 &2 &\lambda &: &\mu \end{bmatrix} \] \[ R_2 \rightarrow R_2 - R_1, \hspace{10pt} R_3 \rightarrow R_3 - R_1 \] \[ \sim \begin{bmatrix} 1 &1 &1 &: &6\\ 0 &1 &2 &: &4\\ 0 &1 &\lambda -1 &: &\mu -6 \end{bmatrix} \] \[ R_3 \rightarrow R_3 - R_2 \] \[ \sim \begin{bmatrix} 1 &1 &1 &: &6\\ 0 &1 &2 &: &4\\ 0 &0 &\lambda -3 &: &\mu -10 \end{bmatrix} \] Case 1: For infinite solutions
\(r(A) = r(A:B) < \text{number of solutions}\), so \(n = 3\)
So \(r(A) = 2 \Rightarrow \lambda - 3 = 0 \)
\( \Rightarrow \boxed{\lambda = 3} \)
\( r(A:B) = 2 \Rightarrow \mu - 10 = 0 \)
\( \Rightarrow \boxed{\mu = 10} \)
Case 2: No solutions
\( r(A) \neq r(A:B) \)
\( r(A) = 2 \Rightarrow \lambda - 3 = 0 \Rightarrow \boxed{\lambda = 3} \)
\( r(A:B) = 3 \Rightarrow \mu - 10 \neq 0 \Rightarrow \boxed{\mu \neq 10} \)
Case 3: Unique solution
\( r(A) = r(A:B) = n = 3 \)
\( \Rightarrow \lambda - 3 \neq 0 \Rightarrow \boxed{\lambda \neq 3} \)
\( \mu \) can have any values -
Eigen Values & Eigen Vectors
To find eigen values and eigen vectors, take a square matrix \(A\) \(\Rightarrow\) find value of [\(A - \lambda I\)], where \(\lambda\) is a scalar quantity \(\Rightarrow\) Put \( |A -\lambda I| = 0 \) i.e. find determinant \(\Rightarrow\) the equation formed is called Characteristic Equation \(\Rightarrow\) on solving, we get values of \(\lambda\) called as Eigen Values \(\Rightarrow\) put (\(A - \lambda I)X = 0\) for each value of \(\lambda\), here \(X\) is called Eigen Vector \(\Rightarrow\) the final equation is \(\boxed{AX = \lambda X}\)
Properties:
(a) Any square matrix \(A\) and its transpose have same eigen values
(b) The sum of eigen values of a matrix is equal to the trace of the matrix (trace is sum of diagonal elements)
(c) The product of eigen values is equal to the determinant of the matrix
Sample Problem 1:
Find Eigen Values and Eigen Vectors - \( \begin{bmatrix} 1 &1 &3\\ 1 &5 &1\\ 3 &1 &1 \end{bmatrix} \)
Solution:
\[ [A - \lambda I] = \begin{bmatrix} 1 - \lambda &1 &3\\ 1 &5 - \lambda &1\\ 3 &1 &1 - \lambda \end{bmatrix} \] \[ \text{Charactristic Equation - } |A - \lambda I| = 0 \] \[ \begin{vmatrix} 1 - \lambda &1 &3\\ 1 &5 - \lambda &1\\ 3 &1 &1 - \lambda \end{vmatrix} = 0 \] \[ (1 - \lambda)[(5 - \lambda)(1 - \lambda) - 1] - 1[(1 - \lambda) - 3] + 3[1 - 3(5 - \lambda)] \] \[ \Rightarrow \lambda^3 -7\lambda^2 - 36 = 0 \] Now, after using Hit & Trial method, we get one value \(\lambda = -2\), so for the other two values, we divide the equation by \(\lambda + 2\), after which we get a quadratic equation: \( \lambda^2 - 9\lambda + 18 \)
\[ (\lambda + 2)(\lambda^2 - 9\lambda + 18) = 0 \] \[ \boxed{\lambda = -2, 6, 3} \] Eigen Vector, \( (A - \lambda I)X = 0 \) \[ \begin{bmatrix} 1 - \lambda &1 &3\\ 1 &5 - \lambda &1\\ 3 &1 &1 - \lambda \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] For \(\lambda = -2\) : \[ \begin{bmatrix} 1-(-2) &1 &3\\ 1 &5-(-2) &1\\ 3 &1 &1-(-2) \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{bmatrix} 3 &1 &3\\ 1 &7 &1\\ 3 &1 &3 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{align*} 3x + y + 3z &= 0\\ x + 7y + z &= 0\\ 3x + y + 3z &= 0 \end{align*} \] By Cross-Multiplication method:
\[ \frac{x}{\begin{vmatrix} 1 &3\\ 7 &1 \end{vmatrix}} = \frac{y}{\begin{vmatrix} 3 &3\\ 1 &1 \end{vmatrix}} = \frac{z}{\begin{vmatrix} 3 &1\\ 1 &7 \end{vmatrix}} \] \[ \frac{x}{-20} = \frac{y}{0} = \frac{z}{20} \] \[ \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \] Eigen Vector - \(< 1, 0, -1 >\)
For \(\lambda = 6\) :
\[ \begin{bmatrix} -5 &1 &3\\ 1 &-1 &1\\ 3 &1 &-5 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{align*} -5x + y + 3z &= 0\\ x - y + z &= 0\\ 3x + y -5z &= 0 \end{align*} \] By Cross-Multiplication method:
\[ \frac{x}{\begin{vmatrix} 1 &3\\ -1 &1 \end{vmatrix}} = \frac{y}{\begin{vmatrix} 3 &-5\\ 1 &1 \end{vmatrix}} = \frac{z}{\begin{vmatrix} -5 &1\\ 1 &-1 \end{vmatrix}} \] \[ \frac{x}{4} = \frac{y}{8} = \frac{z}{4} \] \[ \frac{x}{1} = \frac{y}{2} = \frac{z}{1} \] Eigen Vector - \(< 1, 2, 1 >\)
For \(\lambda = 3\) : \[ \begin{bmatrix} -2 &1 &3\\ 1 &2 &1\\ 3 &1 &-2 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{align*} -2x + y + 3z &= 0\\ x + 2y + z &= 0\\ 3x + y -2z &= 0 \end{align*} \] By Cross-Multiplication method:
\[ \frac{x}{\begin{vmatrix} 1 &2\\ 2 &1 \end{vmatrix}} = \frac{y}{\begin{vmatrix} 3 &-2\\ 1 &1 \end{vmatrix}} = \frac{z}{\begin{vmatrix} -2 &1\\ 1 &2 \end{vmatrix}} \] \[ \frac{x}{-5} = \frac{y}{5} = \frac{z}{-5} \] \[ \frac{x}{1} = \frac{y}{-1} = \frac{z}{1} \] Eigen Vector - \(< 1, -1, 1 >\)
So, \[ \begin{align*} \lambda &= -2 \hspace{10pt} \Rightarrow \hspace{10pt} < 1, 0, -1 >\\ \lambda &= 3 \hspace{10pt} \Rightarrow \hspace{10pt} < 1, -1, 1 >\\ \lambda &= 6 \hspace{10pt} \Rightarrow \hspace{10pt} < 1, 2, 1 >\\ \end{align*} \]
Sample Problem 2:
Find Eigen Vectors and Eigen Vectors for matrix \( \begin{bmatrix} 6 &-2 &2\\ -2 &3 &-1\\ 2 &-1 &3 \end{bmatrix} \)
Solution:
\[ [A - \lambda I] = \begin{bmatrix} 6 - \lambda &-2 &2\\ -2 &3 - \lambda &-1\\ 2 &-1 &3 - \lambda \end{bmatrix} \] \[ \text{Charactristic Equation - } |A - \lambda I| = 0 \] \[ \begin{vmatrix} 6 - \lambda &-2 &2\\ -2 &3 - \lambda &-1\\ 2 &-1 &3 - \lambda \end{vmatrix} = 0 \] \[ (6 - \lambda)[(3 - \lambda)(3 - \lambda) - 1] + 2[-2(3 - \lambda) + 2] + 2[2 - 2(3 - \lambda)] \] \[ \Rightarrow \lambda ^3 - 12\lambda ^2 + 36\lambda - 32 = 0 \] By Hit & Trial method, we get one value \(\lambda = 2\), so for the other two values, we divide the equation by \(\lambda - 2\), after which we get a quadratic equation: \( \lambda^2 - 10\lambda + 16 \)
\[ (\lambda - 2)(\lambda^2 - 10\lambda + 16) = 0 \] \[ \boxed{\lambda = 2, 2, 8} \] Eigen Vector, \( (A - \lambda I)X = 0 \) \[ \begin{bmatrix} 6 - \lambda &-2 &2\\ -2 &3 - \lambda &-1\\ 2 &-1 &3 - \lambda \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] For \(\lambda = 8\) : \[ \begin{bmatrix} -2 &-2 &2\\ -2 &-5 &-1\\ 2 &-1 &-5 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{align*} -2x - 2y + 2z &= 0\\ -2x - 5y - z &= 0\\ 2x - y - 5z &= 0 \end{align*} \] By Cross-Multiplication method:
\[ \frac{x}{\begin{vmatrix} -2 &2\\ -5 &-1 \end{vmatrix}} = \frac{y}{\begin{vmatrix} 2 &-2\\ -1 &-2 \end{vmatrix}} = \frac{z}{\begin{vmatrix} -2 &-2\\ -2 &-5 \end{vmatrix}} \] \[ \frac{x}{12} = \frac{y}{-6} = \frac{z}{6} \] \[ \frac{x}{2} = \frac{y}{-1} = \frac{z}{1} \] Eigen Vector - \(< 2, -1, 1 >\)
For \(\lambda = 2\) : \[ \begin{bmatrix} 4 &-2 &2\\ -2 &1 &-1\\ 2 &-1 & 5 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{align*} 4x - 2y + 2z &= 0\\ -2x + y - z &= 0\\ 2x - y + z &= 0 \end{align*} \] By Cross-Multiplication method:
\[ \frac{x}{\begin{vmatrix} -2 &2\\ 1 &-1 \end{vmatrix}} = \frac{y}{\begin{vmatrix} 2 &4\\ -1 &-2 \end{vmatrix}} = \frac{z}{\begin{vmatrix} 4 &-2\\ -2 &1 \end{vmatrix}} \] \[ \frac{x}{0} = \frac{y}{0} = \frac{z}{0} \] which is not possible, so we take one equation of the three,
\[ 4x - 2y + 2z = 0 \] Put z = 0 \(\Rightarrow 4x - 2y = 0\) \[ 2x = y \Rightarrow \frac{x}{1} = \frac{y}{2} \] Eigen Vector: \( < 1, 2, 0 > \)
For \(\lambda = 2\) again: \[ 4x - 2y + 2z = 0 \] Put y = 0 \( \Rightarrow 4x + 2z = 0 \) \[ \frac{x}{1} = \frac{y}{2} \] Eigen Vector - \( < 1,0,-2 > \) So, \[ \begin{align*} \lambda &= 2 \hspace{10pt} \Rightarrow \hspace{10pt} < 1, 2, 0 >\\ \lambda &= 2 \hspace{10pt} \Rightarrow \hspace{10pt} < 1, 0, -2 >\\ \lambda &= 8 \hspace{10pt} \Rightarrow \hspace{10pt} < 2, -1, 1 >\\ \end{align*} \] -
Cayley-Hamilton Theorem
Cayley-Hamilton Theorem:
It states that every square matrix satisfies its own Characteristic Equation.
Sample Problem :
For the matrix given \( \begin{bmatrix} 2 &1 &1\\ 0 &1 &0\\ 1 &1 &2 \end{bmatrix} \),
(a) Verify Cayley-Hamilton Theorem
(b) Hence find \(A^{-1}\)
(c) Find \(A^8 - 5A^7 + 7A^6 - 3A^5 + A^4 - 5A^3 + 3A^2 - 2A + I\)
(d) Find \(A^{-2}\)Solution:
Characteristic Equation, \(|A - \lambda I| = 0\)
\[ \begin{vmatrix} 2 - \lambda &1 &1\\ 0 &1 - \lambda &0\\ 1 &1 &2 - \lambda \end{vmatrix} = 0 \] \[ (2 - \lambda)[(1 - \lambda)(2 - \lambda)] - 1[0] + 1[-(1 - \lambda)] \] \[ \Rightarrow \lambda ^3 - 5\lambda ^2 + 7\lambda - 3 = 0 \] (a) \[ A^2 = \begin{bmatrix} 2 &1 &1\\ 0 &1 &0\\ 1 &1 &2 \end{bmatrix} \begin{bmatrix} 2 &1 &1\\ 0 &1 &0\\ 1 &1 &2 \end{bmatrix} = \begin{bmatrix} 5 &4 &4\\ 0 &1 &0\\ 4 &4 &5 \end{bmatrix} \] \[ A^3 = \begin{bmatrix} 5 &4 &4\\ 0 &1 &0\\ 4 &4 &5 \end{bmatrix} \begin{bmatrix} 2 &1 &1\\ 0 &1 &0\\ 1 &1 &2 \end{bmatrix} = \begin{bmatrix} 14 &13 &13\\ 0 &1 &0\\ 13 &13 &14 \end{bmatrix} \] Putting \(A\) in place of \(\lambda\) in the Characteristic Polynomial \[ A^3 - 5A^2 + 7A - 3I \] \[ \begin{bmatrix} 14 &13 &13\\ 0 &1 &0\\ 13 &13 &14 \end{bmatrix} - 5\begin{bmatrix} 5 &4 &4\\ 0 &1 &0\\ 4 &4 &5 \end{bmatrix} + 7\begin{bmatrix} 2 &1 &1\\ 0 &1 &0\\ 1 &1 &2 \end{bmatrix} - 3\begin{bmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{bmatrix} \] \[ =\begin{bmatrix} 0 &0 &0\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix} \] Hence verified
(b) Multiplying \(A^{-1}\) on both sides \[ \begin{align*} A^{-1}A^3 - 5A^{-1}A^2 + 7A^{-1}A - 3A^{-1}I &= 0\\ \Rightarrow A^2 - 5A + 7I &= 3A^{-1} \end{align*} \] \[ A^{-1} = \frac{1}{3} \begin{bmatrix} 5 &4 &4\\ 0 &1 &0\\ 4 &4 &5 \end{bmatrix} - \frac{5}{3} \begin{bmatrix} 2 &1 &1\\ 0 &1 &0\\ 1 &1 &2 \end{bmatrix} + \frac{7}{3} \begin{bmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{bmatrix} \] \[ A^{-1} = \frac{1}{3} \begin{bmatrix} 2 &-1 &-1\\ 0 &3 &0\\ -1 &-1 &2 \end{bmatrix} \] (c) \(A^8 - 5A^7 + 7A^6 - 3A^5 + A^4 - 5A^3 + 3A^2 - 2A + I \)
Dividing the given expression by \( A^3 - 5A^2 + 7A - 3I \) to factorize the expression: \[ (A^3 - 5A^2 + 7A - 3I)(A^3 + A) + A^2 + A + I \] \[ = 0 + A^2 + A + I \] \[ = \begin{bmatrix} 5 &4 &4\\ 0 &1 &0\\ 4 &4 &5 \end{bmatrix} + \begin{bmatrix} 2 &1 &1\\ 0 &1 &0\\ 1 &1 &2 \end{bmatrix} + \begin{bmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{bmatrix} \] \[ \begin{bmatrix} 8 &5 &5\\ 0 &3 &0\\ 5 &5 &8 \end{bmatrix} \] (d) Multiplying \(A^{-2}\) on both sides: \[ A^{-2}A^3 - 5A^{-2}A^2 + 7A^{-2}A - 3A^{-2}I = 0 \] \[ A - 5I + 7A^{-1} - 3A^{-2} = 0 \] \[ A^{-2} = \frac{1}{3} (A - 5I + 7A^{-1}) \] \[ A^{-2} = \frac{1}{9} \begin{bmatrix} 5 &-4 &-4\\ 0 &9 &0\\ -4 &-4 &5 \end{bmatrix} \] -
Diagonilization of Matrices
Diagonilization of a matrix \(A\) is the process of reduction of \(A\) to a Diagonal form \(D\) such that \(D = P^{-1}AP\) where \(P\) is called Modal matrix.
Sample Problem:
Find a matrix \(P\) which transforms the given matrix into diagonal matrix: \( \begin{bmatrix} 6 &-2 &2\\ -2 &3 &-1\\ 2 &-1 &3 \end{bmatrix} \)
Solution:
Characteristic Equation for the matrix is: \[ \begin{vmatrix} 6 - \lambda &-2 &2\\ -2 &3 - \lambda &-1\\ 2 &-1 &3 - \lambda \end{vmatrix} = 0 \] \[ (6-\lambda)[(3-\lambda)^2 - 1] + 2[-2(3-\lambda) + 2] + 2[2 - 2(3-\lambda)] \] \[ \lambda ^3 - 12 \lambda ^2 + 36\lambda - 32 = 0 \] \[ (\lambda - 2)^{2}(\lambda - 8) \Rightarrow \lambda = 2,2,8 \] For \(\lambda = 2\):
\[ \begin{bmatrix} 4 &-2 &2\\ -2 &1 &-1\\ 2 &-1 &1 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{align*} 4x -2y + 2z &= 0\\ -2x + y -z &= 0\\ 2x - y + z &= 0 \end{align*} \] Taking first equation, it is satisfied by, \(x = 0, y = 1, z = 1 \)
So, Eigen vector \(X_1\) = \( \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix} \)
Taking first equation again, it is satisfied by, \(x = 1, y = 3, z = 1 \)
So, Eigen vector \(X_2\) = \( \begin{bmatrix} 1\\ 3\\ 1 \end{bmatrix} \)
For \(\lambda = 8\) \[ \begin{bmatrix} -2 &-2 &2\\ -2 &-5 &-1\\ 2 &-1 &-5 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \] \[ \begin{align*} -2x -2y + 2z &= 0\\ -2x - 5y -z &= 0\\ 2x - y - 5z &= 0 \end{align*} \] By Cross-Multiplication method:
\[ \frac{x}{\begin{vmatrix} -2 &2\\ -5 &-1 \end{vmatrix}} = \frac{y}{\begin{vmatrix} 2 &-2\\ -1 &-2 \end{vmatrix}} = \frac{z}{\begin{vmatrix} -2 &-2\\ -2 &-5 \end{vmatrix}} \] \[ \frac{x}{2} = \frac{y}{-1} = \frac{z}{1} \] So, Eigen vector \(X_3\) = \( \begin{bmatrix} 2\\ -1\\ 1 \end{bmatrix} \)
Therefore, \(P = \begin{bmatrix} 0 &1 &2\\ 1 &3 &-1\\ 1 &1 &1 \end{bmatrix} \), \( P^{-1} = \frac{-1}{6} \begin{bmatrix} 4 &1 &-7\\ -2 &-2 &2\\ -2 &1 &-1 \end{bmatrix} \) (using Cayley-Hamilton Theorem)
Now, \(P^{-1}AP = \) \[ \frac{-1}{6} \begin{bmatrix} 4 &1 &-7\\ -2 &-2 &2\\ -2 &1 &-1 \end{bmatrix} \begin{bmatrix} 6 &-2 &2\\ -2 &3 &-1\\ 2 &-1 &3 \end{bmatrix} \begin{bmatrix} 0 &1 &2\\ 1 &3 &-1\\ 1 &1 &1 \end{bmatrix} \] \[ = \begin{bmatrix} 2 &0 &0\\ 0 &2 &0\\ 0 &0 &8 \end{bmatrix} \]